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Bài toán từ quốc gia Iran

[ Thursday, 03 November, 2011 ]

Bài thi học sinh giỏi toán Iran năm 1993:

Trong hình dưới đây, diện tích của các tam giác AOD, DOC và AOB được cho biết. Tìm diện tích của tam giác OEF dựa theo diện tích của những tam giác đã được biết. (Dành cho học sinh Thuận An tại Việt Nam).

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2378561&sid=258ee391a2ebf123ba8b3573e68613c2#p2378561

Solution

Let (Ω) denote the area of shape Ω, I = OE ∩ DC, J = OF ∩ BC, a = (AOD), b = (DOC), c = (AOB) which are the three areas given by the problem, d = (COJ), e = (BOJ), f = (IDE), g = (ICE), h = (CJF), i = (BJF).

We have OAOC = ab = c(d + e), or d + e = bca (i)
Now note that (EBO)(EDO) = g + (ICO) + bcaf + (IDO) = ca , or
ag + bc + (ICO)a = c(f + (IDO)) (ii)

(EDC)(BDC) = (EAC)(BAC), or f + gb + bca = f + g + a + bc + bca, or
f + g = b(a + b)(a + c)a(c – b) (iii)
gf = (ICO)(IDO) (iv)
and (IDO) + (ICO) = b (v)

Solve four equations (ii), (iii), (iv) and (v) with four unknowns f, g, (ICO) and (IDO). By substituting (ICO) = b – (IDO) into (iv) and (ii) we get
gf = b(IDO) – 1 or (IDO) = bff + g and
ag + ab + bc – (IDO)a = c(f + (IDO)), or

(f + g)(ab + ag + bc – cf) = bf(a + c) (vi)

Now substitute f + g from (iii) into equation (vi) to get
(a + b)(ab + ag + bc) = f(bc + 2ac – ab) (vii)
and by substituting g = b(a + b)(a + c)a(c – b) – f into equation (vii)
f = b(a + b)(a + c)²(c – b)(a² + bc + 2ac) and g = bc(a + b)²(a + c)a(c – b)(a² + bc + 2ac).

fg = a(a + c)c(a + b) . Therefore,
(ICO) = b – bff + g = bfg + 1 = bc(a + b)a² + 2ac + bc.
g(ICO) = (a + b)(a + c)a(c – b) .

We also have d + h + ba = i + ec = d + e + h + i + b a + c (viii)
h + ib + d + e = h + i + d + e + ca + b (ix)

Substitute d + e = bca from (i) into (viii) and simplify to get
a(h + i)ab + bc = a(h + i + c) + bca(a + b), or a²(h+ i)(a + b) = ab(h+ i)(a + c) +
bc(a + b)(a + c), or h+ i = bc(a + b)(a + c)a(a² – bc).

On the other hand, substituting the values of d + e and h + i into equation (viii) gives d+ h = bc(a + b)a² – bc.

Lastly, we also have (IFO)(ICO) = (IFE)g = (IFO) + (IFE)(ICO) + g = (OEF)(ICO) + g,

but (IFO) = (ICO) + d + h, and the area of triangle OEF in terms of areas of these three triangles is
(OEF) = [(ICO) + g]×[(ICO) + d + h]/(ICO) = [1 + g(ICO)]×[(ICO) + d + h] = [1 + (a + b)(a + c)a(c – b)]×bc(a + b)[1a² + 2ac + bc + 1a² – bc].

Finally (OEF) = 2bc(a + b)(a + c)(c – b)(a² – bc).

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